The equation |z - i| = |z - 1|,where i = sqrt | vedclass

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(B) Let $z = x + iy$. Then the equation becomes $|x + iy - i| = |x + iy - 1|$.This simplifies to $|x + i(y - 1)| = |(x - 1) + iy|$.Squaring both sides

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the line through the origin with slope $1$

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(B) Let $z = x + iy$. Then the equation becomes $|x + iy - i| = |x + iy - 1|$.This simplifies to $|x + i(y - 1)| = |(x - 1) + iy|$.Squaring both sides,we get $x^2 + (y - 1)^2 = (x - 1)^2 + y^2$.Expanding the terms: $x^2 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2$.Canceling $x^2, y^2,$ and $1$ from both sides,we get $-2y = -2x$,which simplifies to $y = x$.This is the equation of a line passing through the origin with slope $1$.

The equation $|z - i| = |z - 1|$,where $i = \sqrt{-1}$,represents:

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